1. **Problem Statement:** Find the inverse of the matrix $$A = \begin{bmatrix}1 & 2 & 3 \\ 0 & 4 & 5 \\ 6 & 7 & 8\end{bmatrix}$$. 2. **Formula and Rules:** The inverse of a matrix $A$, denoted $A^{-1}$, exists only if $\det(A) \neq 0$. The inverse is given by $$A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A)$$ where $\mathrm{adj}(A)$ is the adjugate (transpose of the cofactor matrix). 3. **Step 1: Calculate the determinant $\det(A)$:** $$\det(A) = 1 \times \begin{vmatrix}4 & 5 \\ 7 & 8\end{vmatrix} - 2 \times \begin{vmatrix}0 & 5 \\ 6 & 8\end{vmatrix} + 3 \times \begin{vmatrix}0 & 4 \\ 6 & 7\end{vmatrix}$$ Calculate minors: $$= 1 \times (4 \times 8 - 5 \times 7) - 2 \times (0 \times 8 - 5 \times 6) + 3 \times (0 \times 7 - 4 \times 6)$$ $$= 1 \times (32 - 35) - 2 \times (0 - 30) + 3 \times (0 - 24)$$ $$= 1 \times (-3) - 2 \times (-30) + 3 \times (-24)$$ $$= -3 + 60 - 72 = -15$$ Since $\det(A) = -15 \neq 0$, the inverse exists. 4. **Step 2: Find the matrix of cofactors:** Calculate each cofactor $C_{ij} = (-1)^{i+j} M_{ij}$ where $M_{ij}$ is the minor determinant. $$C_{11} = (+1) \times \begin{vmatrix}4 & 5 \\ 7 & 8\end{vmatrix} = 32 - 35 = -3$$ $$C_{12} = (-1) \times \begin{vmatrix}0 & 5 \\ 6 & 8\end{vmatrix} = - (0 \times 8 - 5 \times 6) = -(-30) = 30$$ $$C_{13} = (+1) \times \begin{vmatrix}0 & 4 \\ 6 & 7\end{vmatrix} = 0 \times 7 - 4 \times 6 = -24$$ $$C_{21} = (-1) \times \begin{vmatrix}2 & 3 \\ 7 & 8\end{vmatrix} = - (2 \times 8 - 3 \times 7) = - (16 - 21) = 5$$ $$C_{22} = (+1) \times \begin{vmatrix}1 & 3 \\ 6 & 8\end{vmatrix} = 1 \times 8 - 3 \times 6 = 8 - 18 = -10$$ $$C_{23} = (-1) \times \begin{vmatrix}1 & 2 \\ 6 & 7\end{vmatrix} = - (1 \times 7 - 2 \times 6) = - (7 - 12) = 5$$ $$C_{31} = (+1) \times \begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = 2 \times 5 - 3 \times 4 = 10 - 12 = -2$$ $$C_{32} = (-1) \times \begin{vmatrix}1 & 3 \\ 0 & 5\end{vmatrix} = - (1 \times 5 - 3 \times 0) = -5$$ $$C_{33} = (+1) \times \begin{vmatrix}1 & 2 \\ 0 & 4\end{vmatrix} = 1 \times 4 - 2 \times 0 = 4$$ 5. **Step 3: Form the cofactor matrix and transpose it to get the adjugate:** $$\mathrm{adj}(A) = C^T = \begin{bmatrix}-3 & 5 & -2 \\ 30 & -10 & -5 \\ -24 & 5 & 4\end{bmatrix}$$ 6. **Step 4: Calculate the inverse:** $$A^{-1} = \frac{1}{-15} \begin{bmatrix}-3 & 5 & -2 \\ 30 & -10 & -5 \\ -24 & 5 & 4\end{bmatrix} = \begin{bmatrix}\frac{1}{5} & -\frac{1}{3} & \frac{2}{15} \\ -2 & \frac{2}{3} & \frac{1}{3} \\ \frac{8}{5} & -\frac{1}{3} & -\frac{4}{15}\end{bmatrix}$$ **Final answer:** $$A^{-1} = \begin{bmatrix}\frac{1}{5} & -\frac{1}{3} & \frac{2}{15} \\ -2 & \frac{2}{3} & \frac{1}{3} \\ \frac{8}{5} & -\frac{1}{3} & -\frac{4}{15}\end{bmatrix}$$